# Changing the basis of a transformation

Given $T: \mathbb{R}^2 \to \mathbb{R}^2 : T\begin{bmatrix} x \\ y \end{bmatrix} \to \begin{bmatrix} 2x+y \\ x-3y \end{bmatrix}$ with standard basis $\mathcal{B}$ and basis $$\mathcal{B'}=\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \end{bmatrix} \right\}$$ Find the matrix $[T]_\mathcal{B'}^\mathcal{B'}$ representing the transformation $T$ with respect to the basis $\mathcal{B'}$ by making use of a change of basis matrix. Then calculate $[T]_\mathcal{B'}^\mathcal{B'}$ directly and check that you get the same answer.

Method 1

$T \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \qquad \quad T \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \end{bmatrix}$

$\Rightarrow [T]_\mathcal{B}^\mathcal{B} = \left[\begin{array}{rr} 2 & 1 \\ 1 & -3 \end{array}\right]$

To find the transition matrix $[I]_\mathcal{B}^\mathcal{B'}$ from $\mathcal{B}$ to $\mathcal{B'}$ I row reduced the augmented coefficient matrix $$\left[\begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ 1 & 2 & 0 & 1 \end{array}\right] \rightarrow \left[\begin{array}{cc|rr} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 1 \end{array}\right]^*$$ So $[I]_\mathcal{B}^\mathcal{B'}= \left[\begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right]$ and $\left([I]_\mathcal{B}^\mathcal{B'}\right)^{-1}= \left[\begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array}\right]$

Now, $[T]_\mathcal{B'}^\mathcal{B'}= \left([I]_\mathcal{B}^\mathcal{B'}\right)^{-1} [T]_\mathcal{B}^\mathcal{B} [I]_\mathcal{B}^\mathcal{B'} = \left[\begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{rr} 2 & 1 \\ 1 & -3 \end{array}\right] \left[\begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right] = \left[\begin{array}{rr} 8 & -5 \\ 13 & 9 \end{array}\right]$

Method 2

$T \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} \qquad \quad T \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \end{bmatrix}$

$\Rightarrow [T]_\mathcal{B'}^\mathcal{B'}= \left[\begin{array}{rr} 3 & 4 \\ -2 & -5 \end{array}\right]$

Why are my solutions different?

• I'm not going to accept an answer to this question in the foreseeable future because I don't understand linear algebra well enough to understand the correct answers. – ahorn May 24 '16 at 11:21

$T(1,1)$ and $T(1,2)$ should be expressed in basis $\mathcal B'$, not $\mathcal B$.

Let $\{e_1, e_2\}$ be the canonical basis. Set $u_1=\begin{bmatrix}1\\1 \end{bmatrix}=e_1+e_2$, $\,u_2=\begin{bmatrix} 1\\2\end{bmatrix}=e_1+2e_2$. From these, you deduce: $$e_1=2u_1-u_2,\quad e_2=u_2-u_1.$$

You've proved $\,T(u_1)= 3e_1-2e_2$, $\,T(u_2)=4e_1-5e_2$, whence: $$T(u_1)=8u_1-5u_2,\quad T(u_2)=13u_1-9u_2$$ so that $$T_\mathcal B^\mathcal{B'}=\begin{bmatrix}8&13\\-5&-9 \end{bmatrix}.$$

Incidentally, we proved the change of basis matrix from $\mathcal B'$ to $\mathcal B$ is $\,\begin{bmatrix}2&-1\\-1&1 \end{bmatrix}$.

• Did I calculate my transition matrix in Method 1 correctly? I think that it is wrong, and I should have gotten the transition matrix simply by putting the basis vectors of $\mathcal{B'}$ together to form $\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ – ahorn Apr 27 '15 at 10:18
• The transition matrix I got in Method 1, $[T]_\mathcal{B}^\mathcal{B'}$, is supposed to be used by multiplying it with a vector in $\mathcal{B}$ to get a vector in $\mathcal{B'}$. Thus, it is not the transition matrix but rather the inverse. – ahorn Apr 27 '15 at 10:37
• What exactly did I do wrong in method 2? – ahorn Apr 27 '15 at 10:42
• Nothing is wrong. There's one more step to do. – Bernard Apr 27 '15 at 11:05
• What does $\begin{bmatrix} 3 & 4 \\ -2 & -5 \end{bmatrix}$ represent? – ahorn Apr 27 '15 at 11:10